If you know the slope of a line and the angle between them, can you find the slope of the second line? The two lines intersect at (1,4) and the slope of the first line is 3/5. The second line makes a 45 degree angle with the first clockwise so that the second lines slope must be less but I don't know how to find it
 A: Draw the two lines and let $A$ be the point of intersection between the two lines, $B$ the point of intersection of the first line with $x$-axis and $C$ be that between the second line and the $x$-axis. We know the length of $AB$ and the angles $\angle ABC$ and $\angle BAC$. Applying the cosine (or sine) laws on triangle $ABC$, we can find the angle $\angle BCA$. But, we know that the slope of the second line is just: $\tan(\pi - \angle BCA)$.
A: For simplicity, shift the lines so that they intersect at the origin (or draw a new set of axes centered at $(1, 4)$). You know the slope of the first line, so using trigonometry you can find the angle it makes with the $x$ axis. Once you have this, you will be able to find the angle that the second line makes with the $x$ axis. Finally, you can use trigonometry again to translate this into the slope of the second line.
A: Let $\beta$ and $\alpha$ be the slopes of the two lines, so $\tan\beta=\frac{3}{5}$ and $\tan\alpha=m$, the slope of the second line.
Then $\displaystyle\tan\beta=\frac{3}{5}\implies\tan(\alpha+45^\circ)=\frac{3}{5}\implies \frac{\tan\alpha+1}{1-\tan\alpha}=\frac{3}{5}\implies\frac{m+1}{1-m}=\frac{3}{5}\implies m=-\frac{1}{4}.$
A: https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Certain_linear_fractional_transformations
If $f_\alpha(x)=\dfrac{(\cos\alpha)x-\sin\alpha}{(\sin\alpha)x+\cos\alpha}$ then $f_\alpha\circ f_\beta = f_{\alpha+\beta}$.  If $x$ is the slope of a line, then $f_\alpha(x)$ is the slope of its rotation through the angle $-\alpha$.
A: Where $a$ is the slope of your original line, in this case $\frac{3}{5}$.
First find, $\arctan(a)$, this will give you the angle between the $x$ axis and your original line in radians. Second convert the angle between the lines into radians.  If the angle between the lines is counterclockwise add it to $\arctan(a)$, or if it is clockwise subtract it from $\arctan(a)$. So in your case find $\arctan(\frac{3}{5}) - \frac{\pi}{4}$. This gives the angle between your second line and the $x$ axis. To convert this angle into a slope, take the tangent of the quantity. So in your case that would be $\tan(\arctan(\frac{3}{5}) - \frac{\pi}{4})$. Or in the more general case $\tan(\arctan(a)) + \theta)$ (where $\theta$ is the counterclockwise angle).
